2019-2020年高一第二学期期末检测试卷附参考答案
浙江省宁波市2019-2020学年高一下学期期末考试英语试题及答案

宁波市2019学年第二学期期末考试高一英语试卷选择题部分第一部分听力(共两节,满分30分)做题时,先将答案标在试卷上。
录音内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题纸上。
第一节(共5小题;每小题1.5分,满分7.5分)听下面5段对话。
每段对话后有一个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听完每段对话后,你都有10秒钟的时间来回答有关小题和阅读下一小题。
每段对话仅读一遍。
1. Where does the conversation take place?A. In a restaurant.B. In a supermarket.C. In a kitchen.2. Who will probably take the tree away?A. Steve.B. Maggie.C. Jim.3. How is the weather now?A. Rainy.B. Snowy.C. Windy.4. Why does the man seldom wear the shirt?A. It’s too old.B. It doesn’t fit him.C. It has a dark color.5. What are the speakers mainly talking about?A. A holiday.B. The coast.C. A flight.第二节(共15小题;每小题1.5分,满分22.5分)听下面5段对话或独白。
每段对话或独白后有几个小题,从题中所给的A、B、C三个选项中选出最佳选项,并标在试卷的相应位置。
听每段对话或独白前,你将有时间阅读各个小题,每小题5秒钟;听完后,各小题将给出5秒钟的作答时间。
每段对话或独白读两遍。
听下面一段对话,回答第6和7题。
6. Where are the speakers?A. In a store.B. In the man’s house.C. In a restaurant.7. What is in the refrigerator?A. A cup of green tea.B. Large pieces of ice.C. Bottles of water.听下面一段对话,回答第8和9题。
浙江省宁波市2019-2020学年高一化学下学期期末考试试题【含答案】

浙江省宁波市2019-2020学年高一化学下学期期末考试试题考生须知:1.本卷试题分为第I卷、第II 卷,其中第I卷中第二大题选择题分A 组与B 组(B组试题带*),选择 A 卷学校的同学请做A 组试题,,选择B 卷学校的同学请做B 组试题。
本卷满分 100分,考试时间 90 分钟。
2本卷答题时不得使用计算器,不得使用修正液(涂改液)、修正带。
3.答题时将答案均填在答卷相应题号的位置,不按要求答题或答在草稿纸上无效。
4.可能用到的相对原子质量:H-1 C-12N-14O -16Na-23Mg-24S-32 Cl-35.5 K-39Fe-56Cu-64Ag-108 Ba-137第I 卷(选择题,共 50 分)一、选择题(本大题包括20小题,每小题 2 分,共40 分。
每小题只有一个选项符合题意。
)1.下列物质的化学式可用“Na2CO3”表示的是A.烧碱B. 纯碱C.小苏打D. 熟石灰2.垃圾分类有利说资源回收利用,下列垃圾分类不正确的是选项A B C D垃圾废金属剩饭菜过期药品废塑料瓶垃圾分类可回收物厨余垃圾有害垃圾其他垃圾3.以下仪器中,名称为“蒸馏烧瓶”的是A. B. C. D4.下列分散质粒子直径在 10- 9~10-7 m 的分散系是A.稀盐酸B. AgNO3溶液C酒精溶液 D.淀粉溶液5.反应 4HCl+O 2 2Cl 2+2H 2O 中,氧化产物是A.HCl B.O 2C. Cl 2D. H 2O6.下列物质属于电解质的是A.CO 2 B.食盐水 C.MgD.CH 3COOH7.下列有关化学用语表示正确的是A.钠离子的结构示意图:B 乙烯的比例模型:C 中子数为 18 的氯原子Cl D.苯的结构简式:C 6H 637178.下列说法正确的是A.CH 4 与C 5H 12 互为同系物 B . 14C 与14N 互为同位素C.H 2O 与 D 2O 互为同素异形体 D.与 互为同分异构体9.下列说法正确的是A.二氧化碳可用作镁着火的灭火剂B.工业上可以用电解饱和食盐水制取氯气C.Cl 2 能使湿润的有色布条褪色,说明Cl 2具有漂白性D.我国华为AI 芯片已跻身于全球 AI 芯片榜单前列,该芯片的主要材料是二氧化硅10.五种短周期元素在元素周期表中的位置如下图所示,已知M 原子最外层电子数是其电 子层数的 2 倍,下列说法不正确的是A.原子半径: Z>X>YB.X 和M 都能与Y 形成两种常见的化合物C 氢化物的稳定性: N<M D.工业上可以用X 单质与 ZY 2反应制得Z 单质11.下列反应的离子方程式正确的是A.金属钠与水反应: Na+2H 2O =Na ++2OH - +H 2↑B.MgCl2溶液与过量氨水反应: Mg2++2OH-=Mg(OH)2↓C.氯化铁溶液腐蚀铜片: 2Fe3++Cu=2Fe2++ Cu2+D.硫酸铜溶液与氢氧化钡溶液反应: Ba2++ SO42- =BaSO4↓12下列有关物质分离、提纯的说法正确的是A.利用装置①,分离汽油和煤油B.利用装置②,分离蔗糖和食盐C利用装置③,用溴水除去甲烷中的乙烯D.利用装置④,用乙醇提取碘水中的碘13.下列说法正确的是A.石油裂化的目的是为了提高轻质油的产量和质量B煤的液化、气化和石油的分馏都属于物理变化C.为了有效利用生物质能,可以将植物的秸杆直接燃烧获得沼气D.氢气被称为“绿色能源” ,利用化石燃料燃烧放出的热量使水分解产生氢气,是氢能开发的研究方向14.下列说法正确的是A糖类、蛋白质和油脂都是有机高分子化合物B.利用油脂的皂化反应可以制造肥皂C鸡蛋清的溶液中加入硫酸铜溶液,鸡蛋清因盐析而凝聚D.淀粉水解液加过量氢氧化钠溶液后,加新制氢氧化铜悬浊液可检验是否水解完全15燃料电池是一种高效的供能装置,如图是甲烷燃料电池原理示意图,电池工作时,下列有关说法不正确的是A. a 电极发生氧化反应B.b电极是电源的正极C.电子由a 电极经外电路向 b 电极移动,再经电解质溶液回到 a 电极D. b 极电极反应式为: O2+ 2H2O + 4e-=4OH-16.下列说法正确的是A.氯化钠晶体熔融时需要克服离子键B.石英和干冰都属于原子晶体C.CH4、CO2中所有原子均满足最外层8 电子稳定结构D.KOH和 CaCl2都是含有共价键的离子化合物17.反应 CO(g)+ 2H2(g)=CH3OH(g) 的能量变化如下图所示 ,下列说法正确的是A.由图可知, l mol CH3OH(g)的能量低于 2mol H2 (g) 的能量B.断开1 molH2(g)中的化学键需要吸收209.5 kJ 的能量C.CO(g)+ 2H2(g) = CH3OH ( l)ΔH= -91 kJ•mol-1D. CH3OH (g)=CO(g)+ 2 H2 (g)ΔH = 91 kJ•mo1-118.设N A为阿伏加德罗常数的值,下列说法正确的是A.常温常压下,32g O2和O3的混合气体中含有的氧原子数为 2N AB.l mol OH-中含有的电子数为8 N AC. 4.6 gNa与足量O2完全反应,转移的电子数为0.4 N AD.标准状况下,2.24L 水中含有O—H 键的数目为0.2 N A19.已知:2N 2O5 (g) 4NO2 (g)+ O2(g)ΔH =QkJ•mol-1(Q >0), 一定温度下,向2 L 的恒容密闭容器中通入N2O5 ,实验测得的部分数据如下表:时间/s0500l0001500n(N2O5)/mol10.07.0 5.0 5.0下列说法正确的是A.0~1000s内,用N2O5表示的平均反应速率为0.005mol·L-1•s-1B.从表中数据可知,在 1000s 时,反应恰好达到平衡状态C.反应达到平衡时,容器内压强保持不变D 充分反应后,吸收的热量为 5QkJ 20.下列实验方案能达到实验目的的是选项实验目的实验方案A 比较Cl 和Si 非金属性强弱向硅酸钠溶液中滴加盐酸,观察实验现象B配制100mL1.0 mol•L -1 NaOH 溶液称取 NaOH 固体 4.0 g 放入 100mL 容量瓶中,加水溶解,然后桸释至液面与刻度线相切C测定0.01 mol•L -1 NaClO 溶液的 pH用洁净的玻璃棒蘸取待测液点到湿润的 pH 试纸上,变色后与标准比色卡对照D验证氢氧化钡与氯化铵反应为吸热反应在烧杯中加入一定量的氢氧化钡和氯化铵晶体,用玻璃棒搅拌,使之充分混合, 用手触 摸烧杯外壁二、选择题 A ( 本大题包括 5 小题,每小题 2 分,共 10 分, 每小题只有一 个选项符合题意。
福建省福州第一中学2019-2020学年高一下学期期末考试数学答案

福建省福州第一中学2019-2020学年高一下学期期末考试数学试题参考答案1.D 【思路点拨】根据两条直线垂直,列方程求解即可.【解析】由题:直线12:220,:410l x y l ax y +-=++=相互垂直, 所以240a +=, 解得:2a =-. 故选:D【名师指导】此题考查根据两条直线垂直,求参数的取值,关键在于熟练掌握垂直关系的表达方式,列方程求解.2.D 【解析】试题分析:设公比为,由2580a a +=,得,解得,所以.故选D .3.A 【思路点拨】对选项逐一画出图象,由此判断真假性,从而确定正确选项.【解析】对于A 选项,当//αβ时,画出图象如下图所示,由图可知,m n ⊥,故A 选项正确.对于B 选项,当//αβ时,可能m β⊂,如下图所示,所以B 选项错误.对于CD 选项,当αβ⊥时,可能n ⊂α,//m n 如下图所示,所以CD 选项错误.故选:A【名师指导】本小题主要考查线、面位置有关命题真假性的判断,考查空间想象能力,属于基础题.4.B 【解析】本小题主要考查均值定理.11()12x f x x x==≤x x=,即1x =时取等号.故选B . 5.C【解析】本试题主要考查异面直线所成的角问题,考查空间想象与计算能力.延长B 1A 1到E ,使A 1E =A 1B 1,连结AE ,EC 1,则AE ∥A 1B ,∠EAC 1或其补角即为所求,由已知条件可得△AEC 1为正三角形,∴∠EC 1B 为60,故选C .6.B 【思路点拨】先求出动点P 轨迹方程(圆),再根据两圆位置关系确定PQ 的最大值取法,计算即可得结果.【解析】设(,)P x y ,因为2PA PB =2222(2)2(1)x y x y ++-+22(2)4x y ∴-+=因此PQ 22(22)3+2+3=5+3-+故选:B【名师指导】本题考查动点轨迹方程、根据两圆位置关系求最值,考查数形结合思想方法以及基本化简能力,属中档题.7.D 【思路点拨】如图,BCD △中可得30CBD ∠=︒,再利用正弦定理得802BD =,在ABD △中,由余弦定理,即可得答案;【解析】如图,BCD △中,80CD =,15BDC ∠=︒,12015135BCD ACB DCA ∠=∠+∠=︒+︒=︒,∴30CBD ∠=︒, 由正弦定理得80sin135sin 30BD =︒︒,解得802BD =,ACD △中,80CD =,15DCA ∠=︒,13515150ADC ADB BDC ∠=∠+∠=︒+︒=︒,∴15CAD ∠=︒,∴==80AD CD ,ABD △中,由余弦定理得2222cos AB AD BD AD BD ADB =+-⋅⋅∠2280(802)280802cos135=+-⨯⨯⨯︒ 2805=⨯,∴805AB =,即A ,B 两点间的距离为805.故选:D.【名师指导】本题考查正余弦定理的运用,考查函数与方程思想、转化与化归思想,考查逻辑推理能力、运算求解能力.8.C 【思路点拨】取AC 中点D ,连接,BD PD ,证明BD ⊥平面PAC ,故DPB ∠为PB 与平面PAC 所成的角为30,球心O 在平面ABC 的投影为ABC ∆的外心D ,计算得到答案.【解析】取AC 中点D ,连接,BD PD ,2AB BC ==,则BD AC ⊥.PA ⊥平面ABC ,BD ⊂平面ABC ,故PA BD ⊥.PA AC A =,故BD ⊥平面PAC ,故DPB ∠为PB 与平面PAC 所成的角为30.22PB =,故2BD =,6PD =,22AC =,故2ABC π∠=.球心O 在平面ABC 的投影为ABC ∆的外心D , 根据OA OP =知,1,,12OH AP AH HP OD AP ⊥===,故2223R OD AD =+=, 故球的表面积为2412R ππ=. 故选:C.【名师指导】本题考查了三棱锥的外接球问题,确定球心O 在平面ABC 的投影为ABC ∆的外心D 是解题的关键,意在考查学生的计算能力和空间想象能力.9.BD 【思路点拨】对每个选项注意检验,要么证明其成立,要么举出反例判定其错误. 【解析】当0x <时,1x x+为负数,所以A 不正确; 若0a b <<,则110b a<<,考虑函数3()f x x =在R 上单调递增, 所以11()()f f a b >,即3311()()a b>,所以B 正确; 若()20x x -<,则02x <<,2log (,1)x ∈-∞,所以C 不正确; 若0a >,0b >,1a b +≤21,0()224a b a b ab ab ++≤<≤= 所以D 正确. 故选:BD【名师指导】此题考查命题真假性的判断,内容丰富,考查的知识面很广,解题中尤其注意必须对每个选项逐一检验,要么证明其成立,要么举出反例,方可确定选项.10.ABD 【思路点拨】先分析公比取值范围,即可判断A ,再根据等比数列性质判断B,最后根据项的性质判断C,D.【解析】若0q <,则67670,00a a a a <>∴<与671a a >矛盾; 若1q ≥,则11a >∴671,1a a >>∴67101a a ->-与67101a a -<-矛盾; 因此01q <<,所以A 正确;667710101a a a a -<∴>>>-,因此2768(,1)0a a a =∈,即B 正确; 因为0n a >,所以n S 单调递增,即n S 的最大值不为7S ,C 错误;因为当7n ≥时,(0,1)n a ∈,当16n ≤≤时,(1,)n a ∈+∞,所以n T 的最大值为6T ,即D 正确; 故选:ABD【名师指导】本题考查等比数列相关性质,考查综合分析判断能力,属中档题.11.AD 【思路点拨】设(,)C x y ,依题意可确定ABC ∆的外心为(0,2)M ,可得出,x y 一个关系式,求出ABC ∆重心坐标,代入欧拉直线方程,又可得出,x y 另一个关系式,解方程组,即可得出结论.【解析】设(,),C x y AB 的垂直平分线为yx =-,ABC ∆的外心为欧拉线方程为20xy -+=与直线yx =-的交点为(1,1)M -,22||||(1)(1)10MC MA x y ∴==∴++-=,①由()4,0A -,()0,4B ,ABC ∆重心为44(,)33x y -+, 代入欧拉线方程20x y -+=,得20x y --=,② 由 ①②可得2,0x y ==或 0,2x y ==-. 故选:AD【名师指导】本题以数学文化为背景,考查圆的性质和三角形重心,属于较难题.12.ABD 【思路点拨】由正方体的对称性可知,平面α分正方体所得两部分的体积相等;依题意可证1BFD E ,1D F BE ,故四边形1BFD E 一定是平行四边形;当,E F 为棱中点时,EF ⊥平面1BB D ,平面1BFD E ⊥平面1BB D ;当F 与A 重合,当E 与1C 重合时1BFD E 的面积有最大值. 【解析】解: 对于A :由正方体的对称性可知,平面α分正方体所得两部分的体积相等,故A 正确;对于B :因为平面1111ABB A CC D D ,平面1BFD E平面11ABB A BF =,平面1BFD E平面111CC D D D E =,1BFD E ∴.同理可证:1D F BE ,故四边形1BFD E 一定是平行四边形,故B 正确; 对于C :当,E F 为棱中点时,EF ⊥平面1BB D ,又因为EF ⊂平面1BFD E , 所以平面1BFD E ⊥平面1BB D ,故C 不正确;对于D :当F 与A 重合,当E 与1C 重合时1BFD E 的面积有最大值,故D 正确. 故选:ABD【名师指导】本题考查正方体的截面的性质, 解题关键是由截面表示出相应的量与相应的关系,考查空间想象力. 13.52【解析】x 2-x 1=4a -(-2a)=6a =15,解得52a = 14.28π;【解析】由三视图知,圆锥底面的直径为4,所以半径为2,高为23所以母线长为4= ,圆柱的底面直径4,半径为2,高为4.所以该组合体的表面积为224+224228ππππ⨯⨯⨯⨯+⨯= .15.()2,3【思路点拨】先由22sin cos 1A B +=得2B A =,然后利用正弦定理得c b a -2cos 1A =+,再由02π,0π3πB A C A =⎧⎨=-⎩<<<<,求出角A 的范围,从而可得cb a -的取值范围.【解析】解:在ABC 中,因为22sin cos 1A B +=,所以cos cos 2B A =,所以2B A =. 由正弦定理及题设得()sin sin sin sin sin sin A B c Cb a B A B A +==--- sin cos 2cos sin 2sin 2sin A A A AA A+=-()22sin 2cos 12sin cos 2sin cos sin A A A AA A A-+=-24cos 12cos 12cos 1A A A -==+-, 由02π,0π3πB AC A =⎧⎨=-⎩<<<<得π03A <<,故1cos 12A <<,所以cb a-的取值范围为()2,3. 故答案为:()2,3【名师指导】本小题考查解三角形等基础知识;考查运算求解能力;考查数学运算、直观想象等核心素养,体现基础性,属于基础题. 16.215-121n - 【思路点拨】根据和项与通项关系得1112n n S S +-=,再根据等差数列定义与通项公式求1nS ,即得结果,最后根据条件3322a S S =-直接求3.a 【解析】111111120202n n n n n n n n na S S S S S S S S ++++++=∴+=∴--=所以11112(1)2121n n n n S S S n =+-=-∴=- 332112225315a S S =-=-⨯⨯=-故答案为:215-,121n - 【名师指导】本题考查和项与通项关系、等差数列定义与通项公式,考查基本分析求解能力,属基础题.17.【思路点拨】(1)先利用向量求D 点坐标,再根据两点式求直线AD 的方程; (2)先利用向量求cos ABC ∠,再根据三角形面积公式求结果. 【解析】(1)在平行四边形ABCD 中,AB DC =,设(,)(3,1)(2,2)5,1,(5,1)D x y x y x y D ∴-=----∴=-=---所以直线AD 的方程为41454210151y x y x ---=∴-+=+-+; (2)(3,1),(4,5)||10,||41BA BC BA BC =-=--∴==cos 10||||BABC ABC BA BC ⋅∴∠===⋅sin ABC ∴∠=因此平行四边形ABCD 的面积为122||||sin 192ABCSBA BC ABC =⨯⨯∠==【名师指导】本题考查直线方程、三角形面积公式应用、向量数量积求夹角,考查综合分析求解能力,属基础题.18.【思路点拨】(1)不论选那个,都先列出关于公差的方程,解出结果代入等差数列通项公式即可;(2)利用裂项相消法求和. 【解析】(1)322153=15=5S a a =∴∴选①21a -为11a -与31a +的等比中项,则22213(1)(1)(1)(51)(51)(51)a a a d d -=-+∴-=--++2+28012d d d d ∴-=>∴=;选②等比数列{}n b 的公比12q =,12b a =,33b a =, 则23311555()24b a d d -==+=⋅∴=1d >∴舍故只能选①,2(2)52(2)=21n a a n d n n =+-=+-+ (2)111111=()(21)(23)22123n n a a n n n n +=-++++ 所以111111111111()()()()2352572212323233(23)n n T n n n n =-+-++-=-=++++ 【名师指导】本题考查等差数列通项公式、裂项相消法求和,考查基本分析求解能力,属基础题.19.【思路点拨】(1)由菱形性质得AC BD ⊥,由等腰三角形中线的性质得PO BD ⊥,再根据面面垂直的判定定理进行证明即可;(2)利用B CDM M BCD V V --=进行转化,先证出OM ⊥平面ABCD ,从而确定出棱锥的高,利用椎体体积公式求得结果.【解析】(1)证明:设BD 交AC 于点O ,连接PO ,在菱形ABCD 中,AC BD ⊥, 又PB PD =,O 是BD 的中点,∴PO BD ⊥,AC PO O =,AC ⊂平面PAC ,PO ⊂平面PAC ,∴BD ⊥平面PAC ,又BD ⊂平面ABCD ,故平面PAC ⊥平面ABCD ; (2)解:连接OM ,M 为PC 的中点,且O为AC 的中点,∴//OM PA ,由(1)知,BD PA ⊥,又PA AC ⊥, 则BD OM ⊥,OM AC ⊥, 又AC BD O =,∴OM ⊥平面ABCD , 又11122BCDSBD OC =⋅=⨯=132OM PA ==, ∴1133133B CDM M BCD BCDV V SOM --==⋅=⨯⨯=.∴三棱锥B CDM -的体积为1.【名师指导】本题主要考查面面垂直的判定定理以及三棱锥体积的求法. 证明面面垂直,可根据判断定理进行证明,即先由线线垂直证明线面垂直,再由线面垂直证明面面垂直,本质上是证明线面垂直;求三棱锥体积时,如果不能直接求解或者直接求解比较麻烦,可以进行转化,比如本题中,三棱锥B CDM -的体积可以转化为以三角形BCD 为底,求M BCD -的体积.20.【解析】(I )在三角形中,∵1cos 3B =,∴22sin B =. 在ABD ∆中,由正弦定理得sin sin AB AD ADB B=∠,又2AB =,4ADB π∠=,22sin B =.∴83AD =.(II )∵2BD DC =,∴2ABD ADC S S ∆∆=,,又423ADC S ∆=,∴42ABC S ∆= ∵1·sin 2ABC S AB BC ABC ∆=∠,∴6BC =, ∵1·sin 2ABD S AB AD BAD ∆=∠,1·sin 2ADC S AC AD CAD ∆=∠, 2ABD ADC S S ∆∆=,∴sin 2?sin BAD ACCAD AB∠=∠,在ABC ∆中,由余弦定理得2222?cos AC AB BC AB BC ABC =+-∠. ∴42AC =∴sin 2?42sin BAD ACCAD AB∠==∠21.(1)3(2)8【思路点拨】(1)根据等差数列求和公式得n 年每台充电桩总维修费用,再列利润,令利润大于零,解得结果;(2)先列年平均利润,再根据基本不等式求最值.【解析】(1)每台充电桩第n 年总利润为16400[1000(1)400]128002n n n n -+-- 216400[1000(1)400]128000286402n n n n n n -+-->∴-+< 14233142332625.4325n .n n N n ∴-<<+∴<<∈∴≤≤所以每台充电桩第3年开始获利 (2)每台充电桩前n 年的年平均利润16400[1000(1)400]128002n n n n n-+-- ][6464=200282002822400n n n n ⎡⎤⎛⎫-+≤-⋅=⎢⎥ ⎪⎝⎭⎣⎦ 当且仅当64,8n n n==时取等号 所以每台充电桩前8年的年平均利润最大【名师指导】本题考查等差数列实际应用、基本不等式求最值,考查基本分析求解能力,属中档题.22.【解析】(1)由(x ﹣4)2+(y ﹣2)2=20,令x=0,解得y=0或4.∵圆C 2过O ,A 两点,∴可设圆C 2的圆心C 1(a ,2).直线C 2O 的方程为:y=x ,即x ﹣2y=0.∵直线C 2O 与圆C 1相切,∴=,解得a=﹣1,∴圆C 2的方程为:(x+1)2+(y ﹣2)2=,化为:x 2+y 2+2x ﹣4y=0. (2)存在,且为P (3,4).设直线OM 的方程为:y=kx .代入圆C 2的方程可得:(1+k 2)x 2+(2﹣4k )x=0.x M =,y M =.代入圆C 1的方程可得:(1+k 2)x 2﹣(8+4k )x=0.x N=,y N=.设P(x,y),线段MN的中点E.则×k=﹣1,化为:k(4﹣y)+(3﹣x)=0,令4﹣y=3﹣x=0,解得x=3,y=4.∴P(3,4)与k无关系.∴在平面内是存在定点P(3,4)使得PM=PN始终成立.点睛:这个题目考查的是直线和圆的位置关系,一般直线和圆的题很多情况下是利用数形结合来解决的,联立的时候较少;在求圆上的点到直线或者定点的距离时,一般是转化为圆心到直线或者圆心到定点的距离,再加减半径,分别得到最大值和最小值;涉及到圆的弦长或者切线长时,经常用到垂径定理和垂径定理.。
河南省洛阳市2019-2020学年高一下学期期末考试+数学(文)答案

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2019-2020学年度第二学期期末调研考试高一英语试题【含答案】

2019-2020学年度第⼆学期期末调研考试⾼⼀英语试题【含答案】2019-2020 学年度第⼆学期期末调研考试⾼⼀英语试题选择题部分第⼀部分听⼒(共两节,满分30分)做题时,先将答案标在试卷上。
录⾳内容结束后,你将有两分钟的时间将试卷上的答案转涂到答题卡上。
第⼀节(共5⼩题;每⼩题1.5分,满分7.5分)听下⾯5段对话。
每段对话后有⼀个⼩题,从题中所给的A、B、C三个选项中选出最佳选项。
听完每段对话后,你都有10秒钟的时间来回答有关⼩题和阅读下⼀⼩题。
每段对话仅读⼀遍。
1.What will Peter do at 10:00 tomorrow?A. Go camping.B. Stay at home.C. Have a meeting.2.How much did the woman pay for the skirt?A. $10.B. $20.C.$40.3.What's the weather like now?A. Cloudy.B. Sunny.C. Rainy.4. When does the supermarket open on the weekend?A.At 6:00 am.B. At8:00 am.C.At 9:00 am.5.What will the woman do for her mother's birthday?A. Buy a gift.B.Throw a party.C.Make a cake.第⼆节(共15⼩题;每⼩题1.5分,满分22.5分)听下⾯5段对话或独⽩。
每段对话或独⽩后有⼏个⼩题,从题中所给A、B、C三个选项中选出最佳选项。
听每段对话或独⽩前,你将有时间阅读各个⼩题,每⼩题5秒钟;听完后,各⼩题将给出5秒钟的作答时间。
每段对话或独⽩读两遍。
听第6段材料,回答第6、7题。
6. Where does the woman want to go?A.The nearest bank.B.The nearest hospital.C.The nearest post office.7. How will the woman go there?A. By bus.B. By car.C. On foot.听第7段材料,回答第8、9题。
高中生物第二册 2019-2020学年高一生物下学期期末测试卷01(新教材必修二)(含答案)

2019-2020学年高一生物下学期期末测试卷01一、选择题(每题2分共60分)1.一对相对性状的遗传实验中,会导致子二代不符合3︰1性状分离比的情况是( )A.显性基因相对于隐性基因为完全显性B.子一代产生的雌配子中2 种类型配子数目相等,雄配子中也相等C.子一代产生的雄配子中2 种类型配子活力有差异,雌配子无差异D.统计时子二代3 种基因型个体的存活率相等答案:C解析:一对相对性状的遗传实验中,若显性基因相对于隐性基因为完全显性,则子一代为杂合子,子二代性状分离比为3︰1,A正确;若子一代雌雄性都产生比例相等的两种配子,则子二代性状分离比为3︰1,B正确;若子一代产生的雄配子中2种类型配子活力有差异,雌配子无差异,则子二代性状分离比不为3︰1,C错误;若统计时,子二代3 种基因型个体的存活率相等,则表现型比例为3︰1,D正确。
2.桃的果实成熟时,果肉与果皮粘连的称为粘皮,不粘连的称为离皮;果肉与果核粘连的称为粘核,不粘连的称为离核。
已知离皮(A)对粘皮(a)为显性,离核(B)对粘核(b)为显性。
现将粘皮、离核的桃(甲)与离皮、粘核的桃(乙)杂交,所产生的子代出现4种表现型。
由此推断,甲、乙两株桃的基因型分别是( ) A.AABB、aabb B.aaBB、AAbb C.aaBB、Aabb D.aaBb、Aabb答案:D解析:根据基因的分离定律和自由组合定律,子代出现4种表现型,亲本基因型为Aa×Aa或Aa×aa,Bb×Bb 或Bb×bb。
据题意,粘皮为aa,粘核为bb,故甲为aaBb,乙为Aabb。
3.调查发现,人群中夫妇双方均表现正常也能生出白化病患儿。
研究表明白化病由一对等位基因控制,下列有关白化病遗传的叙述,错误的是( )A.致病基因是隐性基因B.如果夫妇双方都是携带者,他们生出白化病患儿的概率是1/4C.如果夫妇一方是白化病患者,他们所生表现正常的子女一定是携带者D.白化病患者与表现正常的人结婚,所生子女表现正常的概率是1答案:D解析:白化病属于常染色体隐性遗传病,A正确。
福建省漳州市2019-2020学年高一下学期期末考试语文试题 Word版含答案

漳州市2019-2020学年下学期期末教学质量检测高一语文试题(满分150分,考试时间150分钟)一、现代文阅读(一)实用类文本阅读(阅读下面的文字,完成1~3题。
材料一:中国人的传统聚会,不论在家中还是在餐馆,如果是享用中餐,一般都是采用围桌会食的方式。
这种亲密接触的方式,是中国饮食文化的一个重要传统。
但纵观中国饮食史,会食方式存在的时间也就是1000多年。
分餐制的历史则可上溯到史前时代,它经过了不少于3000年的发展过程。
古代中国人分餐进食,一般都是席地而坐;后来所说的“席”,正是这古老分餐制的一个写照。
西晋灭亡后,生活在北方的匈奴、鲜卑等族陆续进入中原,先后建立了他们的政权,使得中原地区自殷周以来建立的传统习俗、生活秩序、礼仪制度等受到了一次次强烈的冲击。
正是这种大的历史变革,传统席地而坐的跪姿坐式受到更轻松的垂足坐姿的冲击,及至唐代,各种各样的高足坐具已相当流行,垂足而坐已成为标准姿势。
唐代后期,高椅大桌的会食已十分普通,家具的革新直接影响了饮食方式的变化。
(摘编自王仁湘《分餐制在古代中国至少流行了三千年》,《光明日报》2020年4月18日)材料二:据世界卫生组织统计,影响健康的因素中有60与生活方式和行为有关。
“从科学角度来看,合餐极易导致疾病传染。
部分通过唾液、呼吸道、消化道传播的疾病,如流感、结核病、幽门螺杆菌等,只要就餐人中有人感染此类疾病,就有可能导致其他就餐者感染。
”山西省健康管理师协会专家曹思毅说,在我国,许多人由于过分饮食,高热量食物摄入过多,水果、蔬菜摄入过少,导致肥胖、营养缺乏等问题,从而对健康造成极大的危害。
分餐可以根据每人每餐所需的营养,搭配饭菜,保证每餐有适量的维生素、蛋白质、脂肪等,同时控制进食量,保证营养平衡。
(摘编自马黎《“分餐制”,不能再说“不”!》,《山西日报》2020年2月14日)材料三:新冠肺炎疫情暴发以来,公筷制、分餐制等倡议再次进入公众视野,就目前各地的推行情况来看,需要加强防范“三个忘了”的举措。
2019-2020学年山西省太原市高一下学期期末数学试卷 (解析版)

2019-2020学年山西省太原市高一第二学期期末数学试卷一、选择题(共12小题).1.在等差数列{a n}中,a1=1,d=2,则a4=()A.5B.7C.8D.162.不等式x(x﹣1)>0的解集是()A.(﹣∞,0)B.(0,1)C.(1,+∞)D.(﹣∞,0)∪(1,+∞)3.已知向量=(2,1),=(﹣1,k),⊥,则实数k的值为()A.2B.﹣2C.1D.﹣14.在△ABC中,A=30°,b=,c=1,则a=()A.2B.C.D.15.已知a<b,则下列结论正确的是()A.a2<b2B.<1C.>D.2a<2b6.在等比数列{a n}中,若a1a3a5=8,则a2a4=()A.2B.4C.±2D.±47.cos45°cos15°+sin15°sin45°的值为()A.B.C.D.8.若||=1,||=2,且,的夹角为120°,则|+|的值()A.1B.C.D.29.在数列{a n}中,a1=0,a n+1=(n∈N*),则a2020=()A.0B.C.﹣D.10.已知x>0,y>0,且x+2y=1,则+的最小值是()A.+1B.3+2C.﹣1D.3﹣211.若不等式ax2+2ax﹣1<0对于一切实数x都恒成立,则实数a的取值范围是()A.(﹣∞,﹣1]B.(﹣1,0)C.(﹣1,0]D.[0,+∞)12.已知等差数列{a n}满足a1>0,a2019+a2020>0,a2019•a2020<0.其前n项和为S n,则使S n>0成立时n最大值为()A.2020B.2019C.4040D.4038二、填空题:本大题共4个小题,每个小题3分,共12分,把答案填在横线上.13.已知扇形的半径为1,圆心角为45°,则该扇形的弧长为.14.一船以每小时15km的速度向东航行,船在A处看到一个灯塔B在北偏东60°处;行驶4h后,船到达C处,看到这个灯塔在北偏东15°处.这时船与灯塔的距离为km.15.已知a,b,c成等比数列,a,x,b成等差数列,b,y,c也成等差数列,则+的值为.16.已知数列{a n}满足a n+1+(﹣1)n a n=2n﹣l(n∈N*),则该数列的前80项和为.三、解答题(共3小题,满分30分)17.已知等差数列{a n}中,a2=3,a4=7.等比数列{b n}满足b1=a1,b4=a14.(1)求数列{a n}通项公式a n;(2)求数列{b n}的前n项和S n.18.已知sinα=,α∈(,π).(1)求cosα,tanα;(2)求的值.19.已知△ABC中,A=60°,a=6,B=45°.(1)求b;(2)求△ABC的面积.(请同学们在甲,乙两题中任选一题作答)20.已知向量=(1,cos x),=(1+sin x,1),x∈R,函数f(x)=•﹣1,(1)求函数f(x)的最小正周期和对称中心;(2)若f(x)≥1,求x的取值范围.选做题21.已知向量=(1,cos2x),=(1+sin2x,1),x∈R,函数f(x)=•.(1)求函数f(x)的最小正周期和对称中心;(2)若f(x)≤2,求x的取值范围.(请同学们在甲、乙两题中任选一题作答)22.已知数列{a n}满足a1=3,(n+2)a n+1=(n+3)a n+n2+5n+6(n∈N*).(1)证明:{}为等差数列;(2)设b n=(n∈N*),求数列{b n}的前n项和S n.选做题23.已知数列{a n}满足a1=5,a n+1=2a n+2n+1﹣1(n∈N*),b n=(n∈N*).(1)是否存在实数λ,使得{b n}为等差数列?若存在,求出λ的值;若不存在,请说明理由.(2)利用(1)的结论,求数列{a n}的前n项和S n.参考答案一、选择题:本题共12小题,每小题3分,共36分,在每小题给出的四个选项中,只有一项是符合题目要求的,请将其字母标号填入下表相应位置.1.在等差数列{a n}中,a1=1,d=2,则a4=()A.5B.7C.8D.16【分析】由已知直接利用等差数列的通项公式求解.解:在等差数列{a n}中,由a1=1,d=2,得a4=a1+3d=1+3×2=7.故选:B.2.不等式x(x﹣1)>0的解集是()A.(﹣∞,0)B.(0,1)C.(1,+∞)D.(﹣∞,0)∪(1,+∞)【分析】可以先求出方程x(x﹣1)=0的根,根据一元二次不等式的解法,进行求解;解:x(x﹣1)=0,可得x=1或0,不等式x(x﹣1)>0,解得{x|x>1或x<0},故选:D.3.已知向量=(2,1),=(﹣1,k),⊥,则实数k的值为()A.2B.﹣2C.1D.﹣1【分析】根据条件便有,进行向量数量积的坐标运算便可得出k的值.解:∵;∴;∴k=2.故选:A.4.在△ABC中,A=30°,b=,c=1,则a=()A.2B.C.D.1【分析】利用余弦定理即可求出a的值.解:因为A=30°,b=,c=1,∴a2=b2+c2﹣2bc cos A==1,故a=1.故选:D.5.已知a<b,则下列结论正确的是()A.a2<b2B.<1C.>D.2a<2b【分析】通过举例利用排除法可得ABC不正确,即可得出结论.解:由a<b,取a=﹣2,b=﹣1,可知A,B不正确;取a=﹣1,b=1,可得C不正确.故选:D.6.在等比数列{a n}中,若a1a3a5=8,则a2a4=()A.2B.4C.±2D.±4【分析】根据等比数列的性质知:a1a3a5=(a2q)3=8,a2q=a3=2,a2a4=a32=4.解:设等比数列{a n}的公比为q,则a1a3a5=•a2q•a2q3=(a2q)3=8,则a2q=a3=2.又a2a4=•a3q=a32=22=4.故选:B.7.cos45°cos15°+sin15°sin45°的值为()A.B.C.D.【分析】直接利用两角差的余弦公式,求得所给式子的值.解:cos45°cos15°+sin15°sin45°=(cos45°﹣15°)=cos30°=,故选:B.8.若||=1,||=2,且,的夹角为120°,则|+|的值()A.1B.C.D.2【分析】根据向量的平方等于模的平方,利用数量积定义和数量积的性质即可得出.解:∵||=1,||=2,且,的夹角为120°,∴=1,=4,•=﹣1,∴|+|2=(+)2=+﹣2•=1+4﹣2=3,故|+|=,故选:B.9.在数列{a n}中,a1=0,a n+1=(n∈N*),则a2020=()A.0B.C.﹣D.【分析】利用数列{a n}的通项公式求出数列{a n}的前4项,得到{a n}是周期为3的周期数列,从而a2020=a1,由此能求出结果.解:在数列{a n}中,a1=0,a n+1=(n∈N*),∴=,=﹣,=0,∴{a n}是周期为3的周期数列,∵2020=673×3+1,∴a2020=a1=0.故选:A.10.已知x>0,y>0,且x+2y=1,则+的最小值是()A.+1B.3+2C.﹣1D.3﹣2【分析】利用“乘1法”与基本不等式的性质即可得出.解:因为x>0,y>0,且x+2y=1,则+=(+)(x+2y)=3+,当且仅当且x+2y=1即y==,x=时取等号,故选:B.11.若不等式ax2+2ax﹣1<0对于一切实数x都恒成立,则实数a的取值范围是()A.(﹣∞,﹣1]B.(﹣1,0)C.(﹣1,0]D.[0,+∞)【分析】由已知对a进行分类讨论,然后结合二次不等式的性质可求.解:当a=0时,﹣1<0恒成立,当a≠0时,可得,解可得,﹣1<a<0,综上可得,﹣1<a≤0,故选:C.12.已知等差数列{a n}满足a1>0,a2019+a2020>0,a2019•a2020<0.其前n项和为S n,则使S n>0成立时n最大值为()A.2020B.2019C.4040D.4038【分析】差数列{a n}的首项a1>0,a2019+a2020>0,a2019•a2020<0,可得a2019>0,a2020<0.再利用求和公式及其性质即可得出..解:∵等差数列{a n}的首项a1>0,a2019+a2020>0,a2019•a2020<0,∴a2019>0,a2020<0.于是S4038==>0,S4039==4039•a2020<0.∴使S n>0成立的最大正整数n是4038.故选:D.二、填空题:本大题共4个小题,每个小题3分,共12分,把答案填在横线上.13.已知扇形的半径为1,圆心角为45°,则该扇形的弧长为.【分析】根据弧长公式进行计算即可.解:由题意得,扇形的半径为8cm,圆心角为45°,故此扇形的弧长为:=.故答案为:.14.一船以每小时15km的速度向东航行,船在A处看到一个灯塔B在北偏东60°处;行驶4h后,船到达C处,看到这个灯塔在北偏东15°处.这时船与灯塔的距离为30 km.【分析】根据题意画出相应的图形,求出∠B与∠BAC的度数,再由AC的长,利用正弦定理即可求出BC的长.解:根据题意画出图形,如图所示,可得出∠B=75°﹣30°=45°,在△ABC中,根据正弦定理得:=,即=,∴BC=30km,则这时船与灯塔的距离为30km.故答案为:3015.已知a,b,c成等比数列,a,x,b成等差数列,b,y,c也成等差数列,则+的值为2.【分析】由题意可得b2=ac,2x=a+b,2y=b+c,代入要求的式子+,化简求得结果.解:∵已知a,b,c成等比数列,a,x,b成等差数列,b,y,c也成等差数列,可得b2=ac,2x=a+b,2y=b+c,∴+=+===2,故答案为2.16.已知数列{a n}满足a n+1+(﹣1)n a n=2n﹣l(n∈N*),则该数列的前80项和为3240.【分析】由数列递推式判断数列的特征,4项一组,求和后得到一个等差数列,然后求和即可.解:设a1=a,由a n+1+(﹣1)n a n=2n﹣l,得a2=a+1,a3=2﹣a,a4=7﹣a,a5=a,a6=a+9,a7=2﹣a,a8=15﹣a,a9=a,a10=a+17,a11=2﹣a,a12=23﹣a.可知:a1+a2+a3+a4=10,a5+a6+a7+a8=26,a9+a10+a11+a12=42,…10,26,42,…是等差数列,公差为16,∴数列{a n}的前80项和为:20×10+×16=3240.故答案为:3240.三、解答题(共3小题,满分30分)17.已知等差数列{a n}中,a2=3,a4=7.等比数列{b n}满足b1=a1,b4=a14.(1)求数列{a n}通项公式a n;(2)求数列{b n}的前n项和S n.【分析】(1)设等差数列{a n}的公差为d,运用等差数列的通项公式,解方程可得首项和公差,进而得到所求通项公式;(2)设等比数列{b n}的公比为q,运用等比数列的通项公式,解方程可得公比,进而得到所求和.解:(1)设等差数列{a n}的公差为d,由a2=3,a4=7,可得a1+d=3,a1+3d=7,解得a1=1,d=2,则a n=1+2(n﹣1)=2n﹣1,n∈N*;(2)设等比数列{b n}的公比为q,由b1=a1=1,b4=a14=q3=27,解得q=3,数列{b n}的前n项和S n==(3n﹣1).18.已知sinα=,α∈(,π).(1)求cosα,tanα;(2)求的值.【分析】(1)由题意利用同角三角函数的基本关系,求得结果.(2)由题意利用诱导公式,求得结果.解:(1)∴已知sinα=,α∈(,π),∴cosα=﹣=﹣,∴tanα==﹣.(2)==﹣cos2α=﹣.19.已知△ABC中,A=60°,a=6,B=45°.(1)求b;(2)求△ABC的面积.【分析】(1)由已知利用正弦定理可得b的值.(2)由已知利用两角和的正弦函数公式可求sin C的值,进而根据三角形的面积公式即可求解.解:(1)∵△ABC中,A=60°,a=6,B=45°.∴由正弦定理,可得b===2.(2)∵A+B+C=180°,A=60°,B=45°.∴sin C=sin(A+B)=sin A cos B+cos A sin B=+=,∴S△ABC=ab sin C=×=9+3.(请同学们在甲,乙两题中任选一题作答)20.已知向量=(1,cos x),=(1+sin x,1),x∈R,函数f(x)=•﹣1,(1)求函数f(x)的最小正周期和对称中心;(2)若f(x)≥1,求x的取值范围.【分析】(1)写出f(x)解析式,根据正弦函数的周期及对称中心可得答案;(2)条件等价于sin(x+)≥,解之即可解:由题可得f(x)==1+sin x+cos x﹣1=sin(x+),(1)由f(x)解析式可得其最小正周期T=2π,令x+=kπ,则x=kπ﹣,k∈Z,即f(x)的对称中心为(kπ﹣,0),k∈Z;(2)由f(x)≥1得sin(x+)≥,解得2kπ+≤x+≤2kπ+π,k∈Z,则2kπ≤x≤2kπ+,k∈Z,所以x的取值范围为[2kπ,2kπ+](k∈Z).选做题21.已知向量=(1,cos2x),=(1+sin2x,1),x∈R,函数f(x)=•.(1)求函数f(x)的最小正周期和对称中心;(2)若f(x)≤2,求x的取值范围.【分析】(1)根据平面向量数量积的运算得到f(x)解析式,结合正弦函数性质即可得到答案;(2)由f(x)≤2得到sin(2x+)≤,解之即可解:由题得f(x)==1+sin2x+cos2x=1+sin(2x+)(1)则函数f(x)的最小正周期为T==π,令2x+=kπ,解得x=(k∈Z),即函数的对称中心为(,1)(k∈Z);(2)当f(x)≤2时,即1+sin(2x+)≤2,所以sin(2x+)≤,则﹣+2kπ≤2x+≤+2kπ,解得﹣+kπ≤x≤kπ(k∈Z),即x的取值范围是[﹣+kπ,kπ](k∈Z)(请同学们在甲、乙两题中任选一题作答)22.已知数列{a n}满足a1=3,(n+2)a n+1=(n+3)a n+n2+5n+6(n∈N*).(1)证明:{}为等差数列;(2)设b n=(n∈N*),求数列{b n}的前n项和S n.【分析】(1)直接利用定义的应用求出结果.(2)利用(1)的应用求出数列的通项公式,进一步利用裂项相消法在数列求和中的应用求出结果.【解答】证明:(1)数列{a n}满足a1=3,(n+2)a n+1=(n+3)a n+n2+5n+6(n∈N*).整理得:(常数),所以数列{}是以为首项,1为公差的等差数列.解:(2)由(1)得:,解得:a n=n(n+2).所以.所以:==选做题23.已知数列{a n}满足a1=5,a n+1=2a n+2n+1﹣1(n∈N*),b n=(n∈N*).(1)是否存在实数λ,使得{b n}为等差数列?若存在,求出λ的值;若不存在,请说明理由.(2)利用(1)的结论,求数列{a n}的前n项和S n.【分析】(1)由a n+1=2a n+2n+1﹣1,得,然后利用累加法求得数列{a n}的通项公式,再由等差数列的定义求使{b n}为等差数列的λ值;(2)由(1)知,,令{(n+1)•2n}的前n项和为T n,利用错位相减法求得T n,进一步求得数列{a n}的前n项和S n.解:由a n+1=2a n+2n+1﹣1,得,∴,得,,,…(n≥2).累加得:==.∴(n≥2).a1=5适合上式,∴.则b n==.=.若{b n}为等差数列,则λ﹣1=0,即λ=1.故存在实数λ=1,使得{b n}为等差数列;(2)由(1)知,.令{(n+1)•2n}的前n项和为T n,则,.∴=,得.∴数列{a n}的前n项和S n=n•2n+1+n.。
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静安区2006学年高一第二学期期末检测2019-2020年高一第二学期期末检测试卷附参考答案(105分钟完成; 总分:100分)(牛津教材)2007.6第I卷 (共75分)(第I卷试题的答案请做在答题卡上)I. Listening ComprehensionPart A Short Conversations ( 每小题1分,共10分)Directions:In part A, you will hear ten short conversations between two speakers. At the end of each conversation, a question will be asked about what was said. The conversations and the questions will be spoken only once. After you hear a conversation and the question about it, read the four possible answers and decide which one is the best answer to the question you have heard.1. A. She is going to the shopping center.B. She cannot go with the man.C. She will work with the man tonight.D. She will have a physical exam tomorrow2. A. $0.50 B. $1.50 C. $4.50 D.$2.503. A. Interviewer and interviewee.B.Teacher and student.C.Doctor and nurse.D.Boss and secretary.4. A. At a car shop. B. At a garage. C. In a park. D. In a car showroom.5. A. Finding a larger room. B. Selling the old table.C. Buying another bookshelf.D. Rearranging some furniture.6. A. She seldom works. B. She enjoys working at the same place.C. She often changes her jobs.D. She has worked at the job long.7. A. A nice hair style.B. Martin and John’s wedding.C. An old photo.D. An opening ceremony.8. A. He was busy eating.B. He didn’t notice who John was talking about?C. John was too busy to talk.D. John was meeting the new guests.9. A. Her English is very good.B. She speaks English quickly.C. Her spoken English is still not so good.D. She has no time to learn English.10. A. Ben really wants the scholarship.B.No one wants the scholarship.C.Ben is not interested in the scholarship.D.Others like the scholarship more than Ben.Part B Passages ( 每小题1分,共6分)Direction: In part B, you will hear two short passages, and you will be asked three questions on each of the passages. The passages will be read twice, but the questions will be spoken only once. When you hear a question, read the four possible answers on your paper and decide which one would be the best answer to the question you have heard.Questions 11 to 13 are based on the passage you have just heard.11. A. At a bar in New York.B. At a university restaurant.C. At a Top Club of Britain.D. At a club in the United States.12. A. Because Gloria worked hard for the club.B. Because Gloria regretted giving the big tip.C. Because her story made the club well known.D. Because Gloria had no money in the bank.13. A. A lost –and –found check.B. An unexpected sum of money.C. The biggest tip in history.D. A tip from an English businessman.Questions 14 to 16 are based on the passage you have just heard.14. A. 900 million yuan B. 900 billion yuanC. 2.9 million yuanD. 2.9 billion yuan15. A. Drinking hot soup. B. Taking vitamin C pills.C. Doing more exercise.D. Drinking coffee.16. A. It could move muscle function.B. Muscle damage would not happen.C. Muscle soreness would disappear.D. Upper body soreness would hardly be reduced.Part C Longer Conversations (每小题1分,共4分)Directions:In part C, you will hear two longer conversations. The conversation will be read twice. After you hear each conversation, you are required to fill in the blanks with the information you have heard. Write your answers on your answer sheet.Complete the form. Write ONLY ONE WORD for each answer.II. Grammar ( 每小题1分,共20分)Directions: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.21. I am sure that our new headmaster is ______ you will enjoy working with.A. someoneB. anyoneC. no oneD. everyone22. We have plenty of Coke and juice in the fridge. ______ is no need to buy more drinks.A. ItB. ThereC. ThisD. That23. Karaoke, which was invented by the Japanese, _____ both wonders and problems.A. producedB. has been producedC. has producedD. was produced24. Calvin is confident that it will not be long _____he catches up with the classmates.A. sinceB. that C before D. when25. Choose a nice beach, _____ we will have a picnic there.A. orB. butC. andD. yet26. The kid is crying, for he _____ jump over the ditch..A. dares notB. dares not toC. dare not toD. doesn’t dare to27. If you insist on ______ the car when you are drunk, you are looking for trouble.A. driveB. drivingC. to driveD. having driven28. He never imagined himself _____ behind the big desk in such a magnificent office.A. sittingB. to sitC. being sittingD. sat29. She can’t concentrate herself on her research work with so many letters _____.A. to answerB. to be answeredC. being answeredD. to be answering30. A recent survey reveals that thirty-six percent of home Internet users have been spending lesstime ______ TV since they started using the Internet.A. to watchB. watchingC. watchD. to watching31. I feel it a great honor ______ to give a speech here.A. to askB. to have askedC. to be askedD. being asked32. _____ to sunlight for too much time will do harm to one’s skin.A. ExposedB. Having exposedC. Being exposedD. After being exposed33. This is the first time that your grandpa has been to America, ___________?A. isn’t heB. isn’t itC. hasn’t heD. hasn’t it34. _____ surprised me most was that they had finished the work in such a short time.A. ThatB. ItC. WhatD. How35. I feel most angry about the way _____ I’ve been treated.A. whenB. whereC. whichD.that36. He is sure to come _____ he has some urgent business to do.A. butB. ifC. so thatD. unless37. Is this the reason _______at the meeting for his carelessness in his work?A. he explainedB. what he explainedC. how he explainedD. why he explained38. Please give this ticket to anyone _____ you think is interested in folk music.A. whoB. the oneC. whoeverD. whom39. That e-book is no larger than an ordinary book, with a screen _____ you can read novels.A. in whichB. thatC. whereD. of which40.I have lost the key to the room _____ some important papers are kept.A. in whereB. in whichC. of whichD. under whichIII. Vocabulary(每小题1分,共10分)Directions: Beneath each of the following sentences there are four choices marked A, B, C and D. Choose the one answer that best completes the sentence.41. What follows is the list of goals for salt reduction for our _____ dieter.A. anxiousB. strangeC. effectiveD. typical42. The energy _______ by the chain reaction is transformed into heat.A. conveyedB. releasedC. transmittedD. delivered43. English _____ as a common tongue where people speak many different languages.A. servesB. usesC. regardsD. plays44. His _____ in school is beginning to improve.A. problemB. deedsC. mannersD. behaviour45. Although he is a chef, Robert _____ cooks his own meals.A. bitterlyB. skillfullyC. naturallyD. rarely46. The government has _____ public officials from accepting gifts from foreigners.A. preventedB. benefited C, banned D. made47. Her feelings of self-doubt had _____every relationship that she had ever had.A. keptB. damagedC. destroyedD. lasted48. He has a(n) _____ amount of work to finish before Friday.A. enormousB. bigC. variousD. quantity49.His outstanding _____in the competition proved that he was a worthy winner.A. activityB. performanceC. arrangementD. discussion50. I know how you feel, because I have a _____ problem.A. similarB. differentC. badD. sameIII. Cloze (每小题1分,共10分)Directions: For each blank in the following passages there are four words or phrases marked A, B, C and D. Fill in each blank with the word or phrase that best fits the context.One of the most difficult problems a young person faces is deciding what to do. There are people, of course, who from the time they are six years old, _51_ that they want to be doctors or pilots or fire fighters, but the majority of us do not find the time to make a _52_ about career until somebody or something forces us to _53_ the problem._54_ an occupation(职业) takes time, and there are a lot of things you have to think about when you try to decide what you would like to do. You may find that you will have to take _55_ courses to qualify for a particular kind of work, or you may find out that you will need to get some actual work _56_ to gain enough knowledge.Fortunately, there are a lot of people you can _57_ for advice. At most schools there are teachers who are _58_ qualified to advise you and give you detailed information about the job requirements. And you can talk _59_ your ideas with family members and friends. But even if you get other people _60_ in helping you make a decision, self-evaluation is an important part of the decision-making.51. A. appear B. know C. study D. suggest52. A. arrangement B. timetable C. decision D. appointment53. A. face B. raise C. avoid D. deal54. A. Following B. Inquiring C. Choosing D. Searching55. A. new B. special C. common D. strange56. A. interviews B. profits C. hours D. experiences57. A. get around B. pick out C. turn to D. ask about58. A. professionally B. additionally C. gradually D. essentially59. A. over B. in C. for D. through60. A. participated B. present C. involved D. informedIV. Reading Comprehension ( 每小题1分; 共15分)Directions:Read the following passages. Each passage is followed by several questions orunfinished statements. For each of them, there are four choices marked A, B, C and D. Choose the one that fits best according to the information given in the passage you have just read.(A)As an English child, I started school when I was five. I studied at this school when I was eleven and I will leave next year when I am 16. So I will spend 5 years at this school.The school year starts in September and ends in July. I have two weeks holiday in December, two weeks in April and six weeks in July and August. I also have one week’s holiday in February, June and October.In June next year, before I leave this school, I will take exams in 10 subjects. The exams are called GCSE. All children in England have to study English, mathematics, science and a foreign language. They can also choose some other subjects.My uniform is a black blazer, white shirt, a red and yellow tie and grey trousers. The girls wear grey skirts. There are 1,600 pupils in my school.Here is my timetable for this year. School ends 3:15.Monday In English, we started reading the play called “Macbeth” by William Shakespeare --- it is about a king in Scotland.Tuesday In geography, we are still learning about pollution. It is important but I am bored with it!I played in a cricket match in the evening --- we won!Wednesday In history, my favourite subject, we are learning about medicine in the past. Next week we are going to a museum in London to see some old medical things. We will go by train and it will take one hour.Thursday I went to the cinema with Tom after school.Friday I walked to school with Matthew and I gave him a football for his birthday. Sam came to my house after school.61. In a school year, there are ______ weeks holiday in all.A. 10B. 11C. 12D. 1362. What foreign language does Mike choose this year?A. EnglishB. FrenchC. GermanD. Italian63. According to the passage, the writer thinks that ______.A. Matthew will like the basketball as a birthday presentB. the geography lesson about pollution was important, but boringC. the students in England will leave the school without taking the examsD. he will spend 10 years at school before he goes to colleges or universities( B )On a recent flight, Laura was chatting happily with Lisa, the woman in the next seat until the conversation turned to fares(票价). The woman, who bought her ticket two months in advance, paid $109. Laura paid the full fare of$457. She decided that next time she would find out how to spend less money.Here are some ways to save money:Cheap airplane tickets. To fly for less money, you can buy non-refundable plane tickets two or three months before your trip. The cheapest way to fly is as a courier(送快信的人). In return for delivering a package for a courier company, you get a plane ticket that costs as little as one–quarter of the regular fare----or even less if the company needs someone at the last minute. Recently, a courier flew round trip from Los Angeles to Tokyo for$100;a regular ticket cost around$1,800 .Train passes. If you are going to do a lot of traveling by train, a train pass will save you money. Buying a single pass gives you unlimited travel for a period of time. Train passes can be especially useful in India, which has the world’s largest rail system; in Japan, where trains are fast and convenient; and in Europe, where trains go to over 30,000 cities.Hostels. Hostels used to provide cheap accommodations(住宿)for people under the age of 25. Nowadays, hostels don’t have any age requirements. They are not only cheap($8-$17 a night)but a great way to meet people. Hostels are often in interesting places - - - a castle in Germany, a lighthouse in California, a one – room schoolhouse in the wilderness of Australia. And sometimes hostels have swimming pools.64.If a regular ticket costs$2,000, how much will you pay if you fly as a courier?A.$100. B.$400. C.$500. D.$1.000.65.What is the advantage of train passes?A.They are free for tourists.B.They are convenient to carry.C.They can be used all over the world.D.They can be used unlimitedly during a certain time.66.What is this passage about?A.Travel Tips. B.How to save money.C.Different ways of travel. D.How to travel for less.( C )Ask Steveland Morris and he’ll tell you that blindness is not necessarily disabling. Steveland was born prematurely and totally without sight in 1950. He became Stevie Wonder—composer, singer, and pianist. The winner of ten Grammy awards, Stevie is widely acclaimed for his outstanding contributions to the music world.As a child, Stevie learned not to think about the things he could not do, but to concentrateon the things that he could do. His parents encouraged him to join his sighted brothers in as many activities as possible. They also helped him to sharpen his sense of hearing, the sense upon which the visually disabled are so dependent.Because sound was so important to him, Stevie began at an early age to experiment with different kinds of sound. He would bang things together and then imitate the sound with his voice. Often relying on sound for entertainment, he sang, beat on toy drums, played a toy harmonica(口琴),and listened to the radio.Stevie soon graduated from toy instruments to real instruments. He first learned to play the drums. He then mastered the harmonica and the piano. He became a member of the junior church choir and a leading singer. In the evenings and on weekends, Stevie would play different instruments and sing popular rhythm and blues tunes on the front porches of neighbors’ homes.One of Stevie’s sessions was overheard by Ronnie White, a member of a popular singing group call ed The Miracles. Ronnie immediately recognized Stevie’s talent and took him to audition(试演) for Berry Gordy, the president of Hitsville USA, a large recording company now known as Motown. Stevie recorded his first smash hit “Fingertips” in 1962 at age twel ve, and the rest of Stevie’s story is music history.67. Which of the following is NOT true about Stevie’s chidhood?A. Stevie often told people that a blind person was not necessarily disabled.B. Stevie learnt to concentrate on things that he could do.C. Stevie played as often as possible with his brothers, who had normal sight.D. Stevie tried very hard to train his sense of hearing.68. By saying “Stevie soon graduated from toy instruments to real instruments”, the authormeans that .A. Stevie finished his study at a toy instruments schoolB. Stevie began to study in a real instruments schoolC. Stevie gave up all his toy instruments and began to buy many real instrumentsD. Stevie started to play real instruments69. The author mentions all the following facts EXCEPT that .A. Stevie’s neighbors could often enjoy his playing and singingB. it was Ronnie White that recognized Stevie’s talent and led him to a successful careerC. Berry Gordy helped him to set up his own recording companyD. Stevie’s parents played a very important part in training his sense of hearing70. The “Fingertips” .A. recorded Stevie’s musical performance that won him instant fameB. was a record that turned out to be a great successC. carried the message that the blind could work miracles with their fingertipsD. all of the above( D )Directions: Read the text and choose the most suitable heading from this list for each paragraph of the text. There is one extra heading. (请从答题卡第80题开始做)WAYS TO GET ALONG80.Not everybody has to love me or even like me. I don’t necessarily like everybody I know, so why should everybody else like me? I enjoy being loved, but if somebody doesn’t like me, I wil l still be okay and still feel like I am an okay person. I cannot make somebody like me, any more than someone can get me to like him. I don’t need approval all the time. If someone does not approve of me, I will still be okay.81.People who do thing s I don’t like are not necessarily bad people. They should not be punished just because I don’t like what they do or did. There is no reason why other people should be the way I want them to be, and there is no reason why I should be the way somebody else wants me to be. People will be whatever they want to be, and I will be whatever I want to be. I cannot control other people or change them. They are who they are; we all deserve basic respect.82.I will survive if things are different than what I want them to be. I can accept things the way they are, accept people the way they are, and accept myself the way I am. There is no reason to get upset if I can’t change things to fit my idea of how they ought to be. There is no reason why I should have to lik e everything. Even if I don’t like it, I can live with it.83.I don’t need to watch out for things to go wrong. Things usually go just fine, and when they don’t, I can handle it. I don’t have to waste my energy worrying. The sky won’t fall in; things will be okay. Surely I don’t have to be a certain way because of what has happened in the past. Every day is a new day. It is stupid to think I can’t help being the way I am. Thus when something has gone wrong, I can make some change and correct it.84.I can’t solve other people’s problem for them. I don’t have to take on other people’s problem as if they were my own. I don’t need to change other people or fix up their lives. They are capable and can take care of themselves, and can solve their own problems. I can care and be of some help, but I can’t do everything for them.第Ⅱ卷( 共25分)I. Verb-filling (每小题1分,共10分)Directions: Fill in the each blank with the proper form of the given verb.1.Don’t talk in a loud voice. The baby _____ now. (sleep)2. _____ a better view of the stage, we had to change our seats . (get)3.Look! Here _______ the bus. (come)4. The opening ceremony ______ as planned tomorrow morning unless it rains.( hold)5. Don’t let the child who is not old enough_____ swimming alone.( go)6.It is no good _____ to remember only grammatical rules. You need to practicewhat you have learned. (try)7.I’m afraid the book you are looking for _______ yesterday afternoon. (lend)8. ---What do you think made Angela so upset?---_______ her mobile phone. (lose)9. There are three reading –rooms in our new library, ____ about four hundred students in all.( seat)10. The accident is reported ______ on the first Sunday in December. (occur )II. Translation (每小题3分,共15分)Directions: Translate the following sentences into English, using the words or phrasesgiven in the brackets.1.努力学习,你会通过这次考试的。